Unit 3Midterm and Maxterm Expansions
3 steps to design a single output combinational switching circuit
1) Find a switching function that specifies the desired behavior (OR, AND, etc.)
2) Find a simplified algebraic expression
3) Realize the simplified function using available logic elements
Ex 1) Mary watches TV if it is Monday night and she has finished her homework
3 phrases with two variables
F = 1, Mary watches TV F = 0 otherwise
A = 1, it is Monday night A = 0 otherwise
B = 1, she has finished her hw B = 0 otherwise
Because F is true if A and B are both true, we can write
F = A B
Ex 2) The alarm will ring iff the alarm switch is turned on and the door is not closed or it
is after 6 pm and the window is not closed
5 phrases
B’ = 1, the door is not closed
C = 1, it is after 6 pm
D’ = 1, window is not closed
Z = A B' + C D'
Circuit form
4.2 Combinational Logic Design Using a Truth Table
Where A, B, and C represent 1 st , 2
nd , and 3
rd bits respectively of number N
f = 1 for N ≥ 0112 and f = 0 for N < 0112
A B C f f’
0 0 0 0 1
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 1 0
1 0 1 1 0
1 1 0 1 0
1 1 1 1 0
Derive equation for values of f = 1
f = A' B C + A B' C' + A B' C + A B C' + A B C
Simplifying,
A' B C + A B' (C' + C) + A B (C + C') =
A' B C + A B' + A B =
A' B C + A (B' + B) =
Note:
A + BC
f = (A + B + C) (A + B + C') (A + B' + C)
Can also solve by representing f ' = 1 then take the complement to get f f ' = A' B' C ' + A' B' C + A' B C '
A
C
(f ') ' = (A' B' C ')' (A' B' C)' (A' B C ')'
f = (A + B + C) (A + B + C') (A + B' + C)
(x+y)(x+y')=x
x=A, y=B, z= B'+ C
f = A + B (B' + C)
f = A + BC
4.3 Minterm and Maxterm Expansions
Minterm: term of n variables which is a product of n literals in which each variable
appears exactly once in either true or complemented form, but not both.
Row A B C Minterms
0 0 0 0 A'B'C' = m0
1 0 0 1 A'B'C = m1
2 0 1 0 A'BC' = m2
3 0 1 1 A'BC = m3
4 1 0 0 AB'C' = m4
5 1 0 1 AB'C = m5
6 1 1 0 ABC' = m6
7 1 1 1 ABC = m7
Each minterm has a value of 1 for exactly one combination of values of the variables A,
B and C.
If A = B = C = 0 then A' B' C ' = 1 and is designated as m0
f = A' B C + A B' C + A B' C + A B C' + A B C is an example of a function written as a
sum of minterms. It is often referred to as minterm expansion or standard sum of products.
The above equation can be rewritten in m-notation,
f (A, B, C) = m3 + m4 + m5 + m6 + m7
f (A, B, C) = Σ m (3,4,5,6,7)
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Maxterm: term of n variables which is a sum of n literals in which each variable appears
exactly once in its true or complemented form, but not both.
Row A B C Maxterms
0 0 0 0 A+B+C = M0
1 0 0 1 A+B+C' = M1
2 0 1 0 A+B'+C = M2
3 0 1 1 A+B'+C' = M3
4 1 0 0 A'+B+C = M4
5 1 0 1 A'+B+C' = M5
6 1 1 0 A'+B'+C = M6
7 1 1 1 A'+B'+C' = M7
Each maxterm has a value of zero for exactly one combination of values of A, B, and C.
Thus A = B = C = 0 A + B + C = 0 and is designated M0
Note: The maxterm is the complement of the corresponding minterm
f = (A + B + C) (A + B + C' ) (A + B' + C) is an example of a function written as a
product of maxterms. Referred to as maxterm expansion or standard product of sums
Rewritten in M-notation,
Abbreviated
f (A, B, C) = ΠM (0, 1, 2)
In general, switching expressions can be converted to minterm or maxterm expansions
either by using a truth table or algebraically.
Note: Another way to obtain the minterm expansion is to first write the expression
as a sum of products and introduce missing variables in each term by applying
x + x' = 1
Ex) Find minterm expansion of f = (a, b, c, d) = a' (b' + d) + acd'
= a'b' + a'd + acd'
= a'b' (c + c' ) (d + d' ) + a'd (b + b' ) (c + c' ) + acd' (b + b' )
= a'b'cd + a'b'cd' + a'b'c'd + a'b'c'd' + a'bcd + a'bc'd + a'b'cd + a'b'c'd + abcd' +
ab'cd'
= a'b'c'd' + a'b'c'd + a'b'cd' + a'b'cd + a'bc'd + a'b'c'd + abcd' + ab'cd'
Decimal notation,
Maxterms: ones not listed in minterm for n = 4
Alternate method for maxterm: Factor f to obtain product of sums and introduce missing
variables using xx' = 0 then factor again for maxterm
f = a' (b' + d) + acd'
x' z x y
x yz
= (a + b' +d) (a' + c) (a' + d')
= (a+ b' + cc' + d) (a' + bb' + c + dd') (a' + bb' + cc' + d')
= (a+ b' + c + d) (a+ b' + c' + d) (a' + bb' + c + d) (a' + bb' + c + d')
(a' + bb' + c + d') (a' + bb' + c' + d')
= (a+ b' + c + d) (a+ b' + c' + d) (a' + b + c + d) (a' + b' + c + d) (a' + b + c + d')
(a' + b' + c + d') (a' + b + c' + d') (a' + b' + c' + d')
= (a+ b' + c + d) (a+ b' + c' + d) (a' + b + c + d) (a' + b' + c + d) (a' + b + c + d')
0100 0110 1000 1100 1001
(a' + b' + c + d') (a' + b + c' + d') (a' + b' + c' + d')
1101 1011 1111
f = ΠM (4, 6, 8, 9, 11, 12, 13, 15)
Note: in maxterm translation to decimal, primed values equate to a 1 and
nonprimed to a zero
An equation can be proven valid by factoring the minterm expansions of each side and
showing the expansions are the same
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Ex) a'c + b'c' + ab = a'b' + bc + ac'
Left side,
= a'bc + a'b'c + ab'c' + a'b'c' + abc + abc'
011 001 100 000 111 110
= m3 + m1 + m4 + m0 + m7 + m6
Right side,
= a'b'c + a'b'c' + abc + a'bc + abc' + ab'c'
001 000 111 011 110 100
= m1 + m0 + m7 + m3 + m6 + m4
4.4 General Minterm and Maxterm Expansions
A B C F
0 0 0 a0
0 0 1 a1
0 1 0 a2
0 1 1 a3
1 0 0 a4
1 0 1 a5
1 1 0 a6
1 1 1 a7
For a function n, there are 2 n rows of a truth table and 2^2
n possible functions of n
variables.
7
Maxterm expansion for a general function
7
F = (a0 + m0) (a1 + m1) (a2 + m2) … (az + mz) = Π (ai + mi) i = 0
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Generalized
F = Σ aimi = Π (ai + mi) i = 0 i = 0
2n -1 2n-1
F' = Σ ai'mi = Π (ai' + mi) i = 0 i = 0
Given two different minterms of n variable mi + mj at least one variable appears
complemented in one minterm and in uncomplemented in the other.
Therefore if i ≠ j mi mj = 0
Ex) f1 = Σm(0, 2, 3, 5, 9, 11) f2 = Σm(0, 3, 9, 11, 13, 14)
f1 f2 = Σm(0, 3, 9, 11)
4.5 Incompletely Specified Functions
Assume that the output of N1 has no combination of values for w, x, y, and z to cause A,
B, and C to have a value of 001 and 110. Then when designing N2 it is not necessary to
specify values of F for ABD = 001 and 110.
A B C F
0 0 0 1
0 0 1 X
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 X
1 1 1 1
The X’s indicate “don’t cares” because the values could be 0 or 1 since they will never
occur anyway. A function in this format is said to be incompletely specified.
N1 N2
w x
y z
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To realize a function, a value must be specified for the “don’t cares”. It is best to choose
a value that helps to simplify the function.
Example for both X = 0;
F = A'B'C' + A'BC + ABC
= A'B'C' + BC (A' + A)
F = A'B'C' + A'B'C + A'BC + ABC
= A'B'(C' + C) + BC (A' + A)
= A'B' + BC
F = A'B'C' + A'B'C + A'BC + ABC' + ABC
= A'B'(C' + C) + BC (A' + A) + ABC'
= A'B' + B(C +AC') (Y +XY' = X +Y)
= A'B' + B(C +A)
(The second example yields the simplest solution.)
Minterm expansion for incompletely specified functions; where m = minterm & d = don’t
cares
F = Σ m (0, 3, 7) + Σ d (1, 6)
Maxterm expansion for incompletely specified functions; where M = Maxterm & D =
don’t cares
F = Π M (2, 4, 5) Π D (1, 6)
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4.6 Examples of Truth Table Construction Example 1) Design a binary adder using two 1-bit binary numbers
a b Sum
0 0 0 0 (0 + 0 = 0)
0 1 0 1 (0 + 1 = 1)
1 0 0 1 (1 + 0 = 1)
1 1 1 0 (1 + 1 = 2)
Construct a truth table with logic variables A and B and 2-bit sum logic variables x and y
A B X Y
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0
Because numeric zero is represent by logic zero and numeric 1 by logic 1,
X = AB and Y = A'B + AB' = AB
Example 2) Adder of two 2-bit binary numbers to form a 3-bit sum
N1 N2 N3
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Output function:
X (A, B, C, D) = Σ m (7, 10, 11, 13, 14, 15)
Y (A, B, C, D) = Σ m (2, 3, 5, 6, 8, 9, 12, 15)
Z (A, B, C, D) = Σ m (1, 3, 4, 6, 9, 11, 12, 14)
Example 3) Design an error detector for 6-3-1-1 BCD digit. F = 1 iff if the four
digits (A, B, C, D) are an invalid code. (page 21)
A B C D F
0 0 0 0 0
0 0 0 1 0
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 1
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
Output
= A'CD' + ACD' + ABD
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Example 4) Design a circuit for an 8-4-2-1 BCD digit (A, B, C, D) where Z = 1 if
the decimal is exactly divisible by 3.
0, 3, 6 and 9 are divisible by 3 while 10 – 15 are not valid. (See page 21.)
A B C D Z
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 0
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 X
1 0 1 1 X
1 1 0 0 X
1 1 0 1 X
1 1 1 0 X
1 1 1 1 X
Output:
F(A,B,C,D) = Σ m (0,3,6,9) + Σ d (10,11,12,13,14,15)
To simplify, the easiest solution is a Karnaugh Map which is covered in Unit 5
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4.7 Design of Binary Adders and Subtractors Adders:
Design a parallel adder that adds two 4-bit unsigned binary numbers and a carry
input to give a 4-bit sum and carry output.
Option 1: Use a truth table to design the adder … this is very difficult
Option 2: Design a module for a 2-bit adder with a carry (Full Adder).
Then connect four together to form a 4-bit adder.
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Truth Table
Logic Equation:
= X'(Y'Cin + YCin') + X(Y'Cin' + YCin)
= X'(YCin) X(YCin)'
= (X'YCin + XYCin)+ (XY'Cin + XYCin) + (XYCin' + XYCin)
= YCin(X'+ X) + XCin(Y' +Y) + XY(Cin' + Cin)
= YCin+ XCin + XY
Adders for signed numbers: negatives expressed in complement form.
2’s complement – the last carry is discarded
1’s complement – the last carry is end around carry
Overflow – when two positive numbers yield a negative number
– when two negative numbers yield a positive number
Subtractors: Accomplished by adding the complement of the number to be subtracted.
Use either 1’s or 2’s complement.
Another method is to employ a full subtractor
Where xi, yi and bi are inputs; bi+1 and di are outputs (b = borrow, d = difference)
Truth Table
0 0 0 0 0
0 0 1 1 1
0 1 0 1 1
0 1 1 1 0
1 0 0 0 1
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1